Open cover, subcover, compact metric space, and compact subset of a metric space

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with open sets as in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement and balls as in Open ball, closed ball and sphere in a metric space.

• An open cover of $(X,d)$ is a family $\mathcal{U}$ of open subsets of $X$ with $X = \bigcup \mathcal{U}$, where $\bigcup \mathcal{U} = \{\, x \in X : x \in U \text{ for some } U \in \mathcal{U} \,\}$.
• A subcover of $\mathcal{U}$ is a subfamily $\mathcal{V} \subseteq \mathcal{U}$ that is itself an open cover.
• A family $\mathcal{V}$ of sets is finite when $\mathcal{V} = \emptyset$ or there are $n \in \mathbb{N}$ and sets $V_0, \dots, V_n$ with $\mathcal{V} = \{V_0, \dots, V_n\}$; repetitions in the list are allowed and harmless.
• $(X,d)$ is compact when every open cover of it has a finite subcover: for every open cover $\mathcal{U}$, either $X = \emptyset$ and the empty subfamily covers it, or there are $n \in \mathbb{N}$ and $U_0, \dots, U_n \in \mathcal{U}$ with $X = U_0 \cup \dots \cup U_n .$
• A subset $A \subseteq X$ is a compact subset of $X$ when the metric subspace $(A, d_A)$ is a compact metric space, $d_A$ being the restriction of $d$ to $A \times A$ (Isometry, isometric embedding, and the subspace metric on a subset).

Compactness of a subset is defined intrinsically, and only intrinsically. The last clause speaks about the subspace $(A,d_A)$ and its own open sets, not about families of open subsets of the ambient $X$. The two readings do agree, but that is a theorem and not a convention: it is A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, and no item of this library may use the ambient reading without citing it. Taking the intrinsic reading as the definition is what makes "compact" a property of the metric space $(A,d_A)$ alone, so that a set compact in one ambient space is compact in every other one containing it isometrically.

The empty space is compact, since the empty subfamily of any family covers it; this is the reason the clause above is written with the two cases. The one-point space is compact too, and so is every space listed as $\{x_0, \dots, x_n\}$: given a cover, each $x_i$ lies in some member, and finitely many members chosen in this way already cover.

The finiteness convention, and how it is used both ways. "Finite" above is the listing form, matching the finite lists of Finite intersection property. It agrees with the definition of finiteness by equinumerosity with a natural number (Finite, countably infinite, countable, uncountable), and both directions of the agreement are available and are used below:

• A nonempty finite set $F$ in the sense of Finite, countably infinite, countable, uncountable satisfies $F \approx m$ for some $m \ge 1$, and a bijection $m \to F$ is exactly a listing $F = \{a_0, \dots, a_{m-1}\}$.
• Conversely a set listed as $A = \{a_0, \dots, a_n\}$, that is the image of a function $a$ with domain $\sigma(n)$, is finite in the sense of Finite, countably infinite, countable, uncountable: the map sending $x \in A$ to the least $i \le n$ with $a_i = x$ is an injection of $A$ into $\sigma(n)$, so $A$ is equinumerous with a subset of $\mathbb{N}$ bounded above, and such a subset is finite (Every subset of an at most countable set is at most countable).

Neither direction uses a choice principle: the second selects nothing, taking a least index instead.

Why open covers rather than closed ones. Nothing in the definition would break if $\mathcal{U}$ were allowed to consist of arbitrary sets, but the resulting notion would be uninteresting: every space is covered by its singletons, and only a finite space would survive. Openness of the members is what makes the condition a genuine restriction, and it is what A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it has to keep track of when the ambient space changes.

A warning about the word "cover". A family may cover $A \subseteq X$ without being a family of subsets of $A$: the members are open subsets of $X$ and their union merely contains $A$. That is the ambient reading, and it is a different statement from "$\mathcal{U}$ is an open cover of the metric space $(A,d_A)$", whose members are open subsets of $A$. Which of the two is meant is written out everywhere on this page.